Contents
Andhra Pradesh SSC Class 10 Solutions For Maths – Trigonometry (English Medium)
These Solutions are part of AP SSC Class 10 Solutions for Maths. In this article, we have provided Andhra Pradesh SSC Class 10 Solutions For Maths Chapter 11 Trigonometry
Exercise 11.1:
Question 1:
In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA respectively. Then, find out sin A, cos A and tan A.
Solution :
AP SSC 10th Class Textbook Solutions
Question 2:
The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25cm and ∠Q = 900 respectively. Then find, tan P – tan R.
Solution :
Question 3:
In a right angle triangle ABC with right angle at B, in which a = 24units,b = 25 units and ∠BAC = θ. Then, find cos θ and tan θ.
Solution :
Question 4:
If cos A = \(\frac{12}{13}\) then find sin A and tan A.
Solution :
Question 5:
If 3 tan A = 4, then find sin A and cos A.
Solution :
Question 6:
If ∠A and ∠ X are acute angles such that cos A = cos x then show that ∠A = ∠X.
Solution :
Question 7:
Solution :
Question 8:
In a right angle triangle ABC, right angle is at B, if tan A = \(\sqrt{3}\) then find the value of
- sin A cos C + cos A sin C
- cos A cos C – sin A sin C
Solution :
Exercise 11.2:
Question 1:
Solution :
*Note: Answer for the subparts (iv) and (v) given in the text book are incorrect.
Question 2:
Solution :
Question 3:
Evaluate sin 600 cos 300 + sin 300 cos 600. What is the value of sin(600 + 300). What can you conclude ?
Solution :
Question 4:
Is it right to say cos(600 + 300) = cos 600 cos300 – sin 600 sin 300.
Solution :
Question 5:
In right angle triangle ∆PQR, right angle is at Q and PQ = 6 cms ∠RPQ = 600 Determine the lengths of QR and PR.
Solution :
Question 6:
In ∆XYZ, right angle is at Y, Y Z = x, and XZ = 2x then determine ∠YXZ and ∠YZX.
Solution :
Question 7:
Is it right to say that sin (A+ B) = sin A + sin B? Justify your answer.
Solution :
Exercise 11.3:
Question 1:
Solution :
Question 2:
Solution :
Question 3:
If tan2A = cot(A – 180), where 2A is an acute angle. Find the value of A.
Solution :
Given
tan2A=cot(A -18°)
cot(90-2A)=cot(A -18°)
Since 90-2A and A-18° are both acute angles,
Therefore
90-2A= A-18°
3A=108
A=36°
*Note: Answer given in the text book is incorrect.
Question 4:
If tan A = cot B where A and B are acute angles, prove that A+ B = 900
Solution :
Given tan A=cot B
tan (90-A)=cot B
since (90-A) and B are acute angles so,
90° – A=B
A + B=90°
Thus proved.
Question 5:
If A, B and C are interior angles of a triangle ABC, then show that tan \(\left( \frac{A+B}{2} \right)\) = cot \(\frac{C}{2}\)
Solution :
Question 6:
Express sin 750 + cos 650 in terms of trigonometric ratios of angles between 00 and 450
Solution :
We can write sin 75° =cos (90°-75°)= cos 15°
Cos 65˚=sin(90˚-65˚)=sin 25°
Then, sin 75°+cos 65°= cos 15°+ sin 25°
15° and 25° are between 0° and 45°
Exercise 11.4:
Question 1:
Solution :
Question 2:
Solution :
Question 3:
Solution :
Question 4:
Solution :
Question 5:
Solution :
Question 6:
Simplify secA (1 – sinA) (secA + tanA)
Solution :
Question 7:
Prove that (sinA + cosec)2 + (cosA+ secA)2 = 7 + tan2A+ cot2A
Solution :
Question 8:
Simplify (1 – cos θ) (1 + cosθ) (1 + cot2θ)
Solution :
Question 9:
If secθ + tan θ = p, then what is the value of secθ – tanθ?
Solution :
Question 10:
Solution :
Hope given Andhra Pradesh SSC Class 10 Solutions For Maths chapter 11 Trigonometry are helpful to complete your math homework.
If you have any doubts, please comment below. CBSEtuts try to provide online math tutoring for you.